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If $ \ \ \large \sum_{n=0}^{\infty} \frac{1}{n+1} \binom{2n}{n} x^n= \frac{1-\sqrt{1-4x}}{2x} \ $ , then show that $ \ \sum_{n=0}^{\infty} \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} \ $ .

Answer:

I have no idea how to start this type of .

Any idea ?

G Tony Jacobs
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MAS
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  • If you take that series, multiply by x and differentiate term-by-term ... suggested by the form $$\frac{x^n}{n+1}$$ which is almost a known integral. – GEdgar Oct 04 '17 at 14:00

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$ x\sum_{n=0}^{\infty} \frac{1}{n+1} \binom{2n}{n} x^n= \frac{1-\sqrt{1-4x}}{2} \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n+1} \binom{2n}{n} x^{n+1}= \frac{1-\sqrt{1-4x}}{2} $ . Try to take the derivative of both sides.

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